16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
Myostatin is a protein that is produced in mammalian skeletal muscle cells. It circulates in the blood and acts on muscle tissue to slow down further differentiation and growth. In thoroughbred racehorses, a mutation involving the substitution of a single nucleotide has been identified in the MSTN gene which codes for myostatin. At the site of this mutation, the DNA nucleotide has either a cytosine (C) base or a thymine (T) base, giving race horses three possible genotypes for this mutation: CC, CT or TT. At two years of age, racehorses with the MSTN CC genotype have greater muscle mass than those with the TT genotype. Suggest an explanation for this difference. Racehorses that had won races of different distances were tested to determine their MSTN genotype. The results are shown in . genotype CC percentage of winners with each genotype 1.0 1.2 1.4 1.6 1.8 race distance / km 2.6 2.0 2.2 2.4 CT TT With reference to , describe the effect of the MSTN genotype on the ability of racehorses to win races of different lengths. Modern thoroughbred racehorses are the result of many years of artificial selection. Explain: what is meant by artificial selection how genetic tests for the MSTN genotype can help in the selective breeding of racehorses.
9700_s14_qp_42
THEORY
2014
Paper 4, Variant 2
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Deer mice, Peromyscus maniculatus, are small rodents that live in North America. Like all mammals, their blood contains haemoglobin which combines with oxygen in the lungs, and unloads its oxygen in respiring tissues. Deer mice show variation in their genotypes for the genes that code for the α-polypeptide chain of haemoglobin. In most populations of deer mice, the majority of individuals have the genotype A1A1, while a smaller number have the genotype A0A0. In mice with the genotype A1A1, the amino acid at position 64 in the α-polypeptide chain is aspartic acid. In mice with the genotype A0A0, the amino acid at this position is glycine. Suggest how the change from aspartic acid to glycine in the α-polypeptide chain could have been brought about. The genotypes of deer mice from three different populations, each living at a different altitude, were analysed. shows the relative proportions of deer mice with aspartic acid (white areas) and glycine (black areas) at position 64 in the α-polypeptide of their haemoglobin. altitude 4347 m 1158 m 620 m Describe the effect of altitude on the frequency of the haemoglobin alleles in these populations of deer mice. The partial pressure of oxygen is relatively low at high altitudes. Haemoglobin containing glycine at position 64 in the α-polypeptide chain has a higher affinity for oxygen than haemoglobin with aspartic acid at this position. Suggest how natural selection could account for the difference in allele frequency in deer mice living at high altitudes and low altitudes.
9700_s14_qp_43
THEORY
2014
Paper 4, Variant 3
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The fruit fly, Drosophila melanogaster, has eyes, a striped abdomen and wings longer than its abdomen. This is called a ‘wild-type’ fly. Mutation has resulted in many variations of these features. Table 6.1 shows diagrams of a wild-type fly and three other flies, each of which shows one recessive mutation. Table 6.1 H\HV DEGRPHQ ZLQJGHVFULSWLRQ present striped long present black long absent striped long present striped short Using appropriate symbols, complete the genetic diagram below. symbols parental phenotypes with eyes black abdomen X no eyes striped abdomen parental genotypes gametes offspring genotypes offspring phenotypes with eyes black abdomen no eyes black abdomen with eyes striped abdomen no eyes striped abdomen State how you would carry out a test cross. A cross was carried out between a fly heterozygous for striped abdomen and long wings and a fly with a black abdomen and short wings. The results are shown below in Table 6.2. Table 6.2 offspring number striped abdomen long wing black abdomen long wing striped abdomen short wing black abdomen short wing total A chi-squared test (χ2) was carried out on these data. Complete Table 6.3 and calculate the value of χ2. Table 6.3 observed number (O) expected number (E) O − E (O − E)2 (O − E)2 E χ2 = Σ (O − E)2 E Σ = sum of χ2 Table 6.4 shows χ2 values. Table 6.4 degrees of freedom probability 0.50 0.20 0.10 0.05 0.02 0.01 0.001 2.37 4.64 6.25 7.82 9.84 11.34 16.27 Using Table 6.4, explain what conclusions can be made about the results of the χ2 test.
9700_s16_qp_41
THEORY
2016
Paper 4, Variant 1
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The fruit fly, Drosophila melanogaster, has eyes, a striped abdomen and wings longer than its abdomen. This is called a ‘wild-type’ fly. Mutation has resulted in many variations of these features. Table 6.1 shows diagrams of a wild-type fly and three other flies, each of which shows one recessive mutation. Table 6.1 H\HV DEGRPHQ ZLQJGHVFULSWLRQ present striped long present black long absent striped long present striped short Using appropriate symbols, complete the genetic diagram below. symbols parental phenotypes with eyes black abdomen X no eyes striped abdomen parental genotypes gametes offspring genotypes offspring phenotypes with eyes black abdomen no eyes black abdomen with eyes striped abdomen no eyes striped abdomen State how you would carry out a test cross. A cross was carried out between a fly heterozygous for striped abdomen and long wings and a fly with a black abdomen and short wings. The results are shown below in Table 6.2. Table 6.2 offspring number striped abdomen long wing black abdomen long wing striped abdomen short wing black abdomen short wing total A chi-squared test (χ2) was carried out on these data. Complete Table 6.3 and calculate the value of χ2. Table 6.3 observed number (O) expected number (E) O − E (O − E)2 (O − E)2 E χ2 = Σ (O − E)2 E Σ = sum of χ2 Table 6.4 shows χ2 values. Table 6.4 degrees of freedom probability 0.50 0.20 0.10 0.05 0.02 0.01 0.001 2.37 4.64 6.25 7.82 9.84 11.34 16.27 Using Table 6.4, explain what conclusions can be made about the results of the χ2 test.
9700_s16_qp_43
THEORY
2016
Paper 4, Variant 3
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Domestic goats, Capra hircus, show a wide range of coat patterns and colours. One gene involved in coat colour and pattern has multiple alleles. Four of these alleles are: • A, the allele for white, is dominant to all others • Ab , the allele for badgerface (stripes on face) and Ag, the allele for grey, are codominant • a, the allele for black, is recessive to all others. State all the possible genotypes of: a grey badgerface goat a white goat. A cross between a black goat and a white goat produced a white goat. This white offspring was crossed with a grey goat. The genotype of the grey goat was not known. Use genetic diagrams to show all the possible offspring genotypes and phenotypes that could result from the cross between the white offspring and the grey goat. When goat skin cells produce black melanin pigment these events occur: • Melanocyte stimulating hormone (MSH) is secreted into the bloodstream. • MSH binds to the melanocortin 1 receptor on the cell surface membranes of skin cells. • This causes a rise in cyclic AMP within the cell. • The enzyme tyrosinase is activated and begins melanin production. The white allele A codes for a protein that binds irreversibly to the melanocortin 1 receptor. No rise in cyclic AMP occurs, even when MSH is present in the blood. Explain the appearance of goats with genotype aa.
9700_s18_qp_41
THEORY
2018
Paper 4, Variant 1
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The interpupillary distance (IPD) is the distance in millimetres between the centres of the pupils of the eyes. shows how IPD is measured. Interpupillary distance IPD is one example of a characteristic of human facial structure that shows variation. shows the pattern of variation in IPD in a large sample of adults. interpupillary distance (IPD) / mm number of people Name the type of variation shown in . Suggest and explain how genes and the environment contribute to variation in IPD in humans. Individuals with an IPD of 70 mm or more have a mutation in the PAX3 gene that results in less PAX3 protein being made. The normal role of the PAX3 protein is to increase the expression of many other genes involved in embryonic development. These genes affect a range of phenotypic features such as facial structure, hearing and eye colour. State the term that is used to describe a gene, such as PAX3, that controls the expression of other genes and suggest how the PAX3 protein controls the expression of other genes. Describe how microarray analysis could be used to identify the genes switched on by PAX3 in embryonic cells. The chimpanzee, Pan troglodytes, has DNA that is 98.5% similar to humans, including possession of the PAX3 gene. Investigations show that chimpanzees express higher levels of the PAX3 protein during embryonic development than humans. shows a chimpanzee, Pan troglodytes. Suggest how knowledge of the PAX3 gene helps scientists explain how humans and chimpanzees are very different in facial structure, even though they have very similar DNA.
9700_s19_qp_41
THEORY
2019
Paper 4, Variant 1
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Some neurones in the brain produce a neurotransmitter known as dopamine. Parkinson’s disease occurs when the neurones that produce dopamine die. A person with the disease may experience difficulty in coordinating movement, often seen as tremors in different parts of the body. Parkinson’s disease typically occurs in people older than 55 years. Younger people with these symptoms are said to have early onset Parkinson’s disease (EOPD). Recessive mutations in a gene known as PINK1, located on chromosome 1, an autosome, are believed to be one cause of EOPD. A person with this form of EOPD has a homozygous recessive genotype. Draw a genetic diagram of a cross between two individuals who are heterozygous at the PINK1 gene locus. key to symbols used for alleles parental genotypes gametes offspring genotypes ratio of offspring phenotypes PINK1 codes for a protein kinase enzyme that is important in the functioning of mitochondria in neurones. Most recessive PINK1 mutations are base substitutions which lead to the production of a non-functioning protein kinase enzyme. Explain how a base substitution mutation can lead to the production of a non-functioning protein kinase. One rare, dominant mutation of the PINK1 gene codes for a product that inhibits the normal protein kinase. Explain how this mutation causes EOPD in a heterozygote.
9700_s19_qp_41
THEORY
2019
Paper 4, Variant 1
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Steelhead trout, Oncorhynchus mykiss, are fish that live in streams in North America. To increase the number of steelhead trout, captive breeding has occurred since 1992. Fish eggs and sperm are mixed and the young fish grow in large tanks of aerated water for the first year of their lives. Most are then released into the wild, however a few male and female fish are kept to become the parents of the next generation of captive-bred fish. Each tank may hold up to 50 000 fish. The young captive fish are fed processed food. Some young fish are unable to survive these conditions and a proportion die. Death is usually the result of poor wound-healing after accidents due to overcrowding and due to the spread of diseases. Name the expected pattern of variation in wound-healing ability in a population of fish. Name the process that results in improved survival of captive fish in second, third and subsequent generations of captive-bred fish. Suggest and explain three ways in which the tank environment may make the phenotype of a captive fish different from a wild fish. Two groups of fish were bred and grown in the same environment and were then compared to see if they showed differences in gene expression. The two groups of fish came from: • eggs and sperm from parent fish that had always lived in the wild • eggs and sperm from parent fish that had been bred in captivity. Results from microarray analysis showed that the offspring of the wild and captive-bred fish differed in the expression of over 700 genes. Describe how microarray analysis can detect differences in the expression of many genes when comparing two samples, such as the offspring of wild and captive-bred fish. Explain how gene expression is controlled in eukaryotes such as fish. Many of the differences in gene expression between the offspring of the wild and captive-bred fish were in genes coding for proteins involved in the immune response and in genes coding for proteins involved in wound healing. The fish from captive-bred parents expressed these genes to a greater degree. The researchers concluded that the differences were inherited and adaptive. Explain why the researchers concluded that the differences in gene expression between the two groups of fish were: inherited adaptive.
9700_s19_qp_42
THEORY
2019
Paper 4, Variant 2
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Oculocutaneous albinism (OCA) is a type of albinism. There are many different forms of OCA. OCA1A is one form of OCA, caused by a recessive mutation in the autosomal gene, TYR, coding for the enzyme tyrosinase. This enzyme is involved in the biosynthetic pathway that results in the production of melanin, the pigment responsible for the colour of hair, skin and eyes. A person with OCA1A has white hair, very pale skin and pink eye colour. Draw a genetic diagram to show the probability of a child having OCA1A, if both parents are carriers. Use the symbols A and a for the alleles. parental genotypes gametes offspring genotypes offspring phenotypes probability shows the biosynthetic pathway involving tyrosinase. tyrosinase tyrosinase tyrosine DOPA dopaquinone melanin There are a number of different mutations of the TYR gene that can result in an absence of melanin and cause OCA1A. These include: • a missense mutation, caused by a base substitution, is most common • a nonsense mutation, caused by a base substitution, is less common • an insertion mutation, which is extremely rare. A missense mutation results in a complete polypeptide chain that does not fold properly to form the functioning enzyme. A nonsense mutation results in a shortened polypeptide. Explain why a missense mutation results in a different product from a nonsense mutation. Explain how an insertion mutation in TYR can lead to a lack of melanin in a person with OCA1A. Worldwide, 1 in 17 000 people are born with OCA. This compares with 1 in 165 people among the Guna people of Panama. The Guna people of Panama have a small population and mostly live on many small islands off the coast of Panama. Suggest reasons why the Guna population of Panama has a relatively high number of cases of OCA.
9700_s19_qp_42
THEORY
2019
Paper 4, Variant 2
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Questions Discovered
199